What is the answer to 49^x+1=2(7^x) ?

The answer to 49^x+1=2(7^x) is x=0

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49^x+1=2(7^x)

4 9^{x} + 1 = 2 (7^{x})

x = 0

Solution steps

4 9^{x} + 1 = 2 (7^{x})

Apply exponent rules

(7^{x})^{2} + 1 = 2 \cdot 7^{x}

Rewrite the equation with

7^{x} = u

(u)^{2} + 1 = 2 u

Solve

u^{2} + 1 = 2 u : u = 1

u = 1

Substitute back

u = 7^{x},

solve for

x

Solve

7^{x} = 1 : x = 0

x = 0

e^{4 x} = \frac{1}{16}

x, 2^{x} 4^{y} = 128

4 9^{x - 2} = 34 3^{- x - 3}

2^{x + 3} = \frac{1}{16}

n^{3} = 2^{n}